Modify the Matrix

Modify the Matrix - Problem

Array Matrix Easy

Given a 0-indexed m x n integer matrix matrix, create a new 0-indexed matrix called answer. Make answer equal to matrix, then replace each element with the value -1 with the maximum element in its respective column.

Return the matrix answer.

Input & Output

Example 1 — Basic Matrix

$ Input: matrix = [[1,2,-1],[4,-1,6],[7,8,9]]

› Output: [[1,2,9],[4,8,6],[7,8,9]]

💡 Note: Column maximums are [7,8,9]. Replace -1 in position [0,2] with 9, and -1 in position [1,1] with 8.

Example 2 — Multiple -1s in Same Column

$ Input: matrix = [[3,-1],[5,2]]

› Output: [[3,2],[5,2]]

💡 Note: Column maximums are [5,2]. Replace -1 in position [0,1] with maximum of column 1, which is 2.

Example 3 — No -1s

$ Input: matrix = [[1,2,3],[4,5,6]]

› Output: [[1,2,3],[4,5,6]]

💡 Note: No -1 elements to replace, matrix remains unchanged.

Constraints

m == matrix.length
n == matrix[i].length
1 ≤ m, n ≤ 10⁵
1 ≤ m * n ≤ 10⁵
-10⁹ ≤ matrix[i][j] ≤ 10⁹
The input contains at least one non-negative number.

Visualization

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Asked in

a Amazon 15 G Google 12 M Microsoft 8

The key insight is to avoid scanning columns repeatedly by pre-computing column maximums. Best approach is the two-pass optimization which first finds all column maximums, then replaces all -1s in a second pass. Time: O(m×n), Space: O(n).

Common Approaches

✓ Brute Force - Scan Column for Each -1

⏱️ Time: O(m² × n) Space: O(1)

When we encounter a -1, we iterate through the entire column to find the maximum value, then replace the -1 with that maximum. This approach is straightforward but inefficient for matrices with many -1s.

Two-Pass Optimization

⏱️ Time: O(m × n) Space: O(n)

Instead of scanning columns repeatedly, we first compute the maximum value for each column in one pass, then replace all -1s in a second pass. This eliminates redundant column scans.

Brute Force - Scan Column for Each -1 — Algorithm Steps

Iterate through each cell in the matrix
When finding -1, scan the entire column to find maximum
Replace -1 with the column maximum

Visualization

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Step-by-Step Walkthrough

Find -1

Locate a -1 element in the matrix

Scan Column

Iterate through entire column to find maximum

Replace

Replace -1 with the column maximum

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int** solution(int** matrix, int matrixSize, int* matrixColSize, int* returnSize, int** returnColumnSizes) {
    *returnSize = matrixSize;
    *returnColumnSizes = (int*)malloc(matrixSize * sizeof(int));
    
    for (int i = 0; i < matrixSize; i++) {
        (*returnColumnSizes)[i] = matrixColSize[i];
    }
    
    int m = matrixSize, n = matrixColSize[0];
    
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == -1) {
                // Find maximum in column j
                int colMax = INT_MIN;
                for (int k = 0; k < m; k++) {
                    if (matrix[k][j] != -1) {
                        colMax = (matrix[k][j] > colMax) ? matrix[k][j] : colMax;
                    }
                }
                matrix[i][j] = colMax;
            }
        }
    }
    
    return matrix;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Remove newline if present
    int len = strlen(line);
    if (len > 0 && line[len-1] == '\n') {
        line[len-1] = '\0';
    }
    
    // Parse 2D array from JSON format
    int matrix[100][100];
    int* matrixPtr[100];
    int matrixColSize[100];
    int matrixSize = 0;
    
    // Remove outer brackets
    char* start = strchr(line, '[');
    if (start) start++;
    char* end = strrchr(line, ']');
    if (end) *end = '\0';
    
    char* p = start;
    while (p && *p) {
        // Find start of row
        while (*p && *p != '[') p++;
        if (!*p) break;
        
        // Find end of row
        char* rowStart = p;
        int bracketCount = 0;
        char* rowEnd = p;
        while (*rowEnd) {
            if (*rowEnd == '[') bracketCount++;
            else if (*rowEnd == ']') {
                bracketCount--;
                if (bracketCount == 0) break;
            }
            rowEnd++;
        }
        
        if (bracketCount == 0) {
            // Extract row content
            int rowLen = rowEnd - rowStart + 1;
            char rowStr[1000];
            strncpy(rowStr, rowStart, rowLen);
            rowStr[rowLen] = '\0';
            
            // Parse this row
            int rowSize;
            parseArray(rowStr, matrix[matrixSize], &rowSize);
            matrixPtr[matrixSize] = matrix[matrixSize];
            matrixColSize[matrixSize] = rowSize;
            matrixSize++;
            
            p = rowEnd + 1;
        } else {
            break;
        }
    }
    
    int returnSize;
    int* returnColumnSizes;
    
    int** result = solution(matrixPtr, matrixSize, matrixColSize, &returnSize, &returnColumnSizes);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("[");
        for (int j = 0; j < returnColumnSizes[i]; j++) {
            printf("%d", result[i][j]);
            if (j < returnColumnSizes[i] - 1) printf(",");
        }
        printf("]");
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(returnColumnSizes);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m² × n)

For each cell (m×n), we might scan entire column (m) to find max

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables, modifying matrix in-place

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Scan Column for Each -1 — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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